TRI

602

I El

Biit, _ _

*/~R _l 5 + s/ 5

4(1 — » 2 ) = 4 cos. ’18=2 a: + 3 =---[-3 =-^—

\/l — x 2 = cos. 18 =

^ 5 + V 5

2 V 2

Theorems. (The figures, as marked on the Plate, areaccording to the numbers of the Theorems.)

Theorem 1.— The sides of any plane triangle are as thesines of the angles opposite to them.

Figure 1.—Let abc be a triangle; upon b c let fall theperpendicular a d ; from b a and C a take b e and c h,equal to each other ; on b, with the radius b e, describe thearc e f, cutting b c at f ; on c, with the radius C h, describethe arc H i, cutting b C at i; draw e g and h ic parallel toa d, cutting b c at G and k ; then will e g be the sine of thearc E f, or of the angle ebf; and hk will be the sine ofthe arc H I, or of the angle hci: now, let b e and c h becalled radius, denoted by r, and let e g be called the sine ofangle b, and H K the sine of angle C ; then, by similartriangles,

ba : ad :: BEorCH :eg.

Again, by similar triangles,

ad : AC :: hk : HCorBEj

therefore,

ba:ac::hk:eg;

that is,

BA : A C : : sin’ C : sin’ b.

Therefore, the sides of triangles are as the sines of theiropposite angles.

Theorem 2.— In any plane triangle, as the sum of the tioosides is to their difference, so is the tangent of half the sum ofthe opposite angles to the tangent of half their difference.

Figure 2. —Let A b c be a plane triangle ; from b C cut offb d equal to b a, and join a d ; bisect the angle abc by thestraight line b e ; then will B E also bisect ad; let f be thepoint of section, and let b e meet a C in F.; through f drawf G parallel to A c, cutting B c in G ; then by similar triangles,BCE and B G F,

b G : G c : : b f : f e.

But d c is the difference of the sides b a and b c ; andbecause df:fa::dg:gc; but df and fa are equal,whence d g and G C are equal; therefore, d g, or g c, is halfthe difference of the sides; now half the difference of thesides, added to the lesser side, is half the sum of the sides.It is evident, that bad is half the sum of the angles caband b c A, and d a c is half their difference ; now, by con-sidering A f as radius, b f will be the tangent of the anglebaf, or b A d, and fe the tangent of the angle fae, ord a c ; therefore, as b g, half the sum of the two sides, is toG c, half their difference, so is b f, the tangent of the halfsum of the opposite angles, to F E, the tangent of half theirdifference ; and, consequently, as the sum of the two sides isto their difference, so is the tangent of the half sum of theopposite angles, to the tangent of half their difference.

Theorem 3. —In any plane triangle, if a. perpendicular belet fall upon the longest side from the opposite angle ; then, asthe sum of the segments of the longest side, or base, is to thesum of the other two sides, so is the difference of the sides tothe difference of the segments of the base.

Figure 3. —In the triangle abc, let fall the perpendicularae upon b c ; about a as a centre with the distance, a c, ofthe shortest side, describe a circle ; produce b a to meet thecircumference in g ; then will b g be the sum of the twosides, and b f tlieir difference ; also, B E and E C are thesegments of the base, and bd their difference ; now, by the

property of the circle, bg X bf = bc x bd;whence, b c : bg;: b f : bd; f the

that is, as the longest side, b c, is to b g, the sum 0 ^

other two sides, so is b f, the difference of the sides, tothe difference of the segments of the base. _ ff j e ; 9

When two angles of a triangle are given, the third an ^ roinfound by subtracting the sum of the two given angle 9180°, or two right angles; and, consequently, if ° ne 0 - n .

acute angles of a right-angled triangle be given, the re ^ing angle will be found, by subtracting the acute angle ^ ^90 ; for, in this case also, two angles are given, viz-,the acute angles and the right angle ; and these beingtracted from two right angles, or 180, the other acute a Dwill be found. . ar e

Various propositions might be given, but the precedi gsufficient for every case of plane trigonometry. , ce d

The analogies of plane trigonometry may easily be ae _from those of spherical trigonometry ; the former bew 0 rticular cases of the latter. uclJl -

Proposition .— The sum, s, and difference, d, of two 9tides, x and y, being given ; to find the quantities them

Let x + y = s

x — y = d s + d,

add these equations together, and 2 x = s + d, or * = '2

_ s df

subtract these equations from each other, and 2 y —

s d n tiU eS>

or y =—-—; whence x, the greater of the two q ua

2 . d V

is half the sum and difference of these quantities ; an

lesser, is half of the difference between the sum an palf

ence ; or, in other words, the half sum added to gyp-

difference, gives the greater quantity ; and the halt 9tracted from the half difference, gives the lesser q uan \ £SSolutions of the three cases of oblique-angled triangEvery plane triangle consists of six parts, the tn g peand the three angles ; three of these parts must a ‘ .

given, and of these given parts one at least must t)to find the remaining three parts. 0

Case 1 .—Two angles and a side being given, to J 1remaining sides. _ . th 0

As the sine of the angle opposite the given side 19 ^ g [desine of the angle opposite the required side, so is the gto the required side. . anI rle ^

In the oblique-angled triangle, ABC, given59°, the angle c 52° 15', and the side a b 276.5, 0and b c.

We shall find

angle b = 180° — (52° 15' + 59°) = 68° .45, g.89jff.

as the sine C 52° .. "9^59^

is to the sine of L b 68° ..

so is the side A b 276.5 .

'W'

to the side a c 325.9.

To find b c. 9 . 8 -fi®}

As the sine of £ C 52° 15' .

is to the sine of /_ A 59° .

so is A b. 276 5 .

to b c. 299.8 .

Geometrical Construction. , py th 0

5 J tb&

From a scale of equal parts draw a b = * < f ' jfoW .protractor, or line of chords, make angle a o a iig* 0triangle cannot be constructed without havin'*